Chemistry, 01.08.2021 02:30 arias58

Consider the dissociation of strong versus weak acids: HCl(aq)+H2O(l)→H3O+(aq)+Cl−(aq)HF(a q)+H2O(l)⇌H3O+(aq)+F−(aq)

The first reaction is not reversible, but the second one is. So, only the conjugate of the weak acid, F−, can react with H3O+. Now consider the solubility of insoluble salts:

AgCl(s)⇌Ag+(aq)+Cl−(aq)AgF(s)⇌Ag+(a q)+F−(aq)

The addition of acid has no effect on silver chloride. But for the second reaction, H3O+ will react with F−, decreasing its concentration and driving the equilibrium to the right. Thus, salts that contain the conjugate of a weak acid become more soluble as the acidity of the solution increases.

Part A
Identify the effect of increasing acidity on the solubility of the given compounds.
Drag the appropriate items to their respective bins.

Ca3(PO4)2
ZnS
BaCO3
AgI
Mg(OH)2
MgBr2

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the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

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cytokinesis is the splitting of a cell membrane to form 2 identical daughter cells. it's similar to mitosis as both cells have identical copies of dna and cytokinesis is a step in the mitotic cycle.

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Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.

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the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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Consider the dissociation of strong versus weak acids: HCl(aq)+H2O(l)→H3O+(aq)+Cl−(aq)HF(a q)+H2O(...

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